Given : let 
and 
be three students whose chances of solving a problem is given i.e, 
and 
Now, 
(excatly one of them will solve it) 
and 
and 
and 
and not 
and not A and not B)
![\begin{array}{l} =\mathrm{P}(\mathrm{A} \cap \bar{B} \cap \bar{C})+\mathrm{P}(\mathrm{B} \cap \bar{A} \cap \bar{C})+\mathrm{P}(\mathrm{C} \cap \bar{A} \cap \bar{B}) \\ =\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\bar{B}) \times \mathrm{P}(\bar{C})+\mathrm{P}(\mathrm{B}) \times \mathrm{P}(\bar{A}) \times \mathrm{P}(\bar{C})+\mathrm{P}(\mathrm{C}) \times \mathrm{P}(\bar{B}) \times \mathrm{P}(\bar{A}) \\ =\left[\frac{1}{3} \times \frac{3}{4} \times \frac{5}{6}\right]+\left[\frac{1}{4} \times \frac{2}{3} \times \frac{5}{6}\right]+\left[\frac{1}{6} \times \frac{3}{4} \times \frac{2}{3}\right] \\ =\frac{15}{72}+\frac{10}{72}+\frac{6}{72} \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-73e25e2050ecf76f9ef19b972779bd59_l3.png "Rendered by QuickLaTeX.com")
Therefore, The probability that excatly one of them will solve the problem is 
The probabilities of 
solving a problem are 
and 
, respectively. If all the three try to solve the problem simultaneously, find the probability that exactly one of them will solve it.
solving a problem are and , respectively. If all the three try to solve the problem simultaneously, find the probability that exactly one of them will solve it.The probabilities of solving a problem are and , respectively. If all the three try to solve the problem simultaneously, find the probability that exactly one of them will solve it.
solving a problem are and , respectively. If all the three try to solve the problem simultaneously, find the probability that exactly one of them will solve it.